Coincidence? $\left(\frac 1e\right)^{\frac 1e}\approx \ln 2$

Question

Is it a coincidence that $$\color{lightgrey}{0.6922\cdots =}\left(\frac 1e\right)^{\frac 1e}\approx \ln 2\color{lightgrey}{=0.6931\cdots}$$ ?

I don't know anything having to do with a seemingly unconnected property, but for what it's worth, the two real solutions to $x^x = \ln 2$ are, to a $4$-decimal approximation, $0.4000$ and $0.3366,$ and the average of these two values is $0.3683,$ which is fairly close to $\frac{1}{e} = 0.3679$ $(4$-place approximate value). — Dave L. Renfro, Nov 13 '19 at 17:02
In mathematics, as in real life (whatever that means), nothing is ever a coincidence -- at least not when examined sufficiently deeply. — Allawonder, Nov 13 '19 at 17:03
I do not believe this to be of any coincidence, nor significance. For example, $$\bigg(\frac13\bigg)^{\frac13}\approx 0.69336$$ is even closer to $\ln 2$. — Andrew Chin, Nov 13 '19 at 18:07

score 3 · Answer 1 · edited Nov 01 '22 at 07:31

I'm not sure if this helps, but I have found that the Taylor series for these values end up being quite similar.

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\ldots,$$

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\ldots\,.$$

$\ln(2)=\ln(1+1)$ and $(\frac{1}{e})^\frac{1}{e}=e^\frac{-1}{e}$.

Therefore,

$$\ln(2)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\ldots,$$

$$\left(\frac{1}{e}\right)^\frac{1}{e}=1-\frac{1}{e}+\frac{1}{2!e^2}-\frac{1}{3!e^3}+\frac{1}{4!e^4}-\frac{1}{5!e^5}+\ldots\,.$$

1 Answers1